Gaussian elimination
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Gaussian Elimination. Chemical Engineering Majors Author(s): Autar Kaw http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates. Naïve Gauss Elimination http://numericalmethods.eng.usf.edu. Naïve Gaussian Elimination.

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Gaussian Elimination

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Gaussian elimination

Gaussian Elimination

Chemical Engineering Majors

Author(s): Autar Kaw

http://numericalmethods.eng.usf.edu

Transforming Numerical Methods Education for STEM Undergraduates


Na ve gauss elimination http numericalmethods eng usf edu

Naïve Gauss Eliminationhttp://numericalmethods.eng.usf.edu


Na ve gaussian elimination

Naïve Gaussian Elimination

A method to solve simultaneous linear equations of the form [A][X]=[C]

Two steps

1. Forward Elimination

2. Back Substitution


Forward elimination

Forward Elimination

The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix


Forward elimination1

Forward Elimination

A set of n equations and n unknowns

. .

. .

. .

(n-1) steps of forward elimination


Forward elimination2

Forward Elimination

Step 1

For Equation 2, divide Equation 1 by and multiply by .


Forward elimination3

Forward Elimination

Subtract the result from Equation 2.

_________________________________________________

or


Forward elimination4

Forward Elimination

Repeat this procedure for the remaining equations to reduce the set of equations as

. ..

. ..

. ..

End of Step 1


Forward elimination5

Forward Elimination

Step 2

Repeat the same procedure for the 3rd term of Equation 3.

. .

. .

. .

End of Step 2


Forward elimination6

Forward Elimination

At the end of (n-1) Forward Elimination steps, the system of equations will look like

. .

. .

. .

End of Step (n-1)


Matrix form at end of forward elimination

Matrix Form at End of Forward Elimination


Back substitution

Back Substitution

Solve each equation starting from the last equation

Example of a system of 3 equations


Back substitution starting eqns

Back Substitution Starting Eqns

. .

. .

. .


Back substitution1

Back Substitution

Start with the last equation because it has only one unknown


Back substitution2

Back Substitution


Gaussian elimination

THE END

http://numericalmethods.eng.usf.edu


Na ve gauss elimination example http numericalmethods eng usf edu

Naïve Gauss EliminationExamplehttp://numericalmethods.eng.usf.edu


Example liquid liquid extraction

Example: Liquid-Liquid Extraction

A liquid-liquid extraction process conducted in the Electrochemical Materials Laboratory involved the extraction of nickel from the aqueous phase into an organic phase. A typical set of experimental data from the laboratory is given below:

Assuming g is the amount of Ni in organic phase and a is the amount of Ni in the aqueous phase, the quadratic interpolant that estimates g is given by


Example liquid liquid extraction1

Example: Liquid-Liquid Extraction

The solution for the unknowns x1, x2, and x3 is given by

Find the values of x1, x2, and x3 using Naïve Gauss Elimination. Estimate the amount of nickel in organic phase when 2.3 g/l is in the aqueous phase using quadratic interpolation.


Number of steps of forward elimination

Number of Steps of Forward Elimination

Number of steps of forward elimination is

(n-1)=(3-1)=2


Example liquid liquid extraction2

Example: Liquid-Liquid Extraction

Solution

Forward Elimination: Step 1

Yields


Example liquid liquid extraction3

Example: Liquid-Liquid Extraction

Forward Elimination: Step 1

Yields


Example liquid liquid extraction4

Example: Liquid-Liquid Extraction

Forward Elimination: Step 2

Yields

This is now ready for Back Substitution


Example liquid liquid extraction5

Example: Liquid-Liquid Extraction

Back Substitution: Solve for x3 using the third equation


Example liquid liquid extraction6

Example: Liquid-Liquid Extraction

Back Substitution: Solve for x2 using the second equation


Example liquid liquid extraction7

Example: Liquid-Liquid Extraction

Back Substitution:Solve for x1 using the first equation


Example liquid liquid extraction8

Example: Liquid-Liquid Extraction

The solution vector is

The polynomial that passes through the three data points is then

Where g is the amount of nickel in the organic phase and a is the amount of in the aqueous phase.


Example liquid liquid extraction9

Example: Liquid-Liquid Extraction

When 2.3 g/l is in the aqueous phase, using quadratic interpolation, the estimated amount of nickel in the organic phase is


Gaussian elimination

THE END

http://numericalmethods.eng.usf.edu


Na ve gauss elimination pitfalls http numericalmethods eng usf edu

Naïve Gauss EliminationPitfallshttp://numericalmethods.eng.usf.edu


Gaussian elimination

Pitfall#1. Division by zero


Is division by zero an issue here

Is division by zero an issue here?


Is division by zero an issue here yes

Is division by zero an issue here? YES

Division by zero is a possibility at any step of forward elimination


Gaussian elimination

Pitfall#2. Large Round-off Errors

Exact Solution


Gaussian elimination

Pitfall#2. Large Round-off Errors

Solve it on a computer using 6 significant digits with chopping


Gaussian elimination

Pitfall#2. Large Round-off Errors

Solve it on a computer using 5 significant digits with chopping

Is there a way to reduce the round off error?


Avoiding pitfalls

Avoiding Pitfalls

  • Increase the number of significant digits

  • Decreases round-off error

  • Does not avoid division by zero


Avoiding pitfalls1

Avoiding Pitfalls

  • Gaussian Elimination with Partial Pivoting

  • Avoids division by zero

  • Reduces round off error


Gaussian elimination

THE END

http://numericalmethods.eng.usf.edu


Gauss elimination with partial pivoting http numericalmethods eng usf edu

Gauss Elimination with Partial Pivotinghttp://numericalmethods.eng.usf.edu


Pitfalls of na ve gauss elimination

Pitfalls of Naïve Gauss Elimination

  • Possible division by zero

  • Large round-off errors


Avoiding pitfalls2

Avoiding Pitfalls

  • Increase the number of significant digits

  • Decreases round-off error

  • Does not avoid division by zero


Avoiding pitfalls3

Avoiding Pitfalls

  • Gaussian Elimination with Partial Pivoting

  • Avoids division by zero

  • Reduces round off error


What is different about partial pivoting

What is Different About Partial Pivoting?

At the beginning of the kth step of forward elimination, find the maximum of

If the maximum of the values is

in the p th row,

then switch rows p and k.


Matrix form at beginning of 2 nd step of forward elimination

Matrix Form at Beginning of 2nd Step of Forward Elimination


Example 2 nd step of fe

Example (2nd step of FE)

Which two rows would you switch?


Example 2 nd step of fe1

Example (2nd step of FE)

Switched Rows


Gaussian elimination with partial pivoting

Gaussian Elimination with Partial Pivoting

A method to solve simultaneous linear equations of the form [A][X]=[C]

Two steps

1. Forward Elimination

2. Back Substitution


Forward elimination7

Forward Elimination

Same as naïve Gauss elimination method except that we switch rows before each of the (n-1) steps of forward elimination.


Example matrix form at beginning of 2 nd step of forward elimination

Example: Matrix Form at Beginning of 2nd Step of Forward Elimination


Matrix form at end of forward elimination1

Matrix Form at End of Forward Elimination


Back substitution starting eqns1

Back Substitution Starting Eqns

. .

. .

. .


Back substitution3

Back Substitution


Gaussian elimination

THE END

http://numericalmethods.eng.usf.edu


Gauss elimination with partial pivoting example http numericalmethods eng usf edu

Gauss Elimination with Partial PivotingExamplehttp://numericalmethods.eng.usf.edu


Example 2

Example 2

Solve the following set of equations by Gaussian elimination with partial pivoting


Example 2 cont

Example 2 Cont.

  • Forward Elimination

  • Back Substitution


Forward elimination8

Forward Elimination


Number of steps of forward elimination1

Number of Steps of Forward Elimination

Number of steps of forward elimination is (n-1)=(3-1)=2


Forward elimination step 1

Forward Elimination: Step 1

Examine absolute values of first column, first row

and below.

  • Largest absolute value is 144 and exists in row 3.

  • Switch row 1 and row 3.


Forward elimination step 1 cont

Forward Elimination: Step 1 (cont.)

Divide Equation 1 by 144 and

multiply it by 64, .

.

Subtract the result from Equation 2

Substitute new equation for Equation 2


Forward elimination step 1 cont1

Forward Elimination: Step 1 (cont.)

Divide Equation 1 by 144 and

multiply it by 25, .

.

Subtract the result from Equation 3

Substitute new equation for Equation 3


Forward elimination step 2

Forward Elimination: Step 2

Examine absolute values of second column, second row

and below.

  • Largest absolute value is 2.917 and exists in row 3.

  • Switch row 2 and row 3.


Forward elimination step 2 cont

Forward Elimination: Step 2 (cont.)

Divide Equation 2 by 2.917 and

multiply it by 2.667,

.

Subtract the result from Equation 3

Substitute new equation for Equation 3


Back substitution4

Back Substitution


Back substitution5

Back Substitution

Solving for a3


Back substitution cont

Back Substitution (cont.)

Solving for a2


Back substitution cont1

Back Substitution (cont.)

Solving for a1


Gaussian elimination with partial pivoting solution

Gaussian Elimination with Partial Pivoting Solution


Gauss elimination with partial pivoting another example http numericalmethods eng usf edu

Gauss Elimination with Partial PivotingAnother Examplehttp://numericalmethods.eng.usf.edu


Partial pivoting example

Partial Pivoting: Example

Consider the system of equations

In matrix form

=

Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping


Partial pivoting example1

Partial Pivoting: Example

Forward Elimination: Step 1

Examining the values of the first column

|10|, |-3|, and |5| or 10, 3, and 5

The largest absolute value is 10, which means, to follow the rules of Partial Pivoting, we switch row1 with row1.

Performing Forward Elimination


Partial pivoting example2

Partial Pivoting: Example

Forward Elimination: Step 2

Examining the values of the first column

|-0.001| and |2.5| or 0.0001 and 2.5

The largest absolute value is 2.5, so row 2 is switched with row 3

Performing the row swap


Partial pivoting example3

Partial Pivoting: Example

Forward Elimination: Step 2

Performing the Forward Elimination results in:


Partial pivoting example4

Partial Pivoting: Example

Back Substitution

Solving the equations through back substitution


Partial pivoting example5

Partial Pivoting: Example

Compare the calculated and exact solution

The fact that they are equal is coincidence, but it does illustrate the advantage of Partial Pivoting


Gaussian elimination

THE END

http://numericalmethods.eng.usf.edu


Gaussian elimination

Determinant of a Square MatrixUsing Naïve Gauss EliminationExamplehttp://numericalmethods.eng.usf.edu


Theorem of determinants

Theorem of Determinants

If a multiple of one row of [A]nxn is added or subtracted to another row of [A]nxn to result in [B]nxn then det(A)=det(B)


Theorem of determinants1

Theorem of Determinants

The determinant of an upper triangular matrix [A]nxn is given by


Forward elimination of a square matrix

Forward Elimination of a Square Matrix

Using forward elimination to transform [A]nxn to an upper triangular matrix, [U]nxn.


Example

Example

Using naïve Gaussian elimination find the determinant of the following square matrix.


Forward elimination9

Forward Elimination


Forward elimination step 11

Forward Elimination: Step 1

Divide Equation 1 by 25 and

multiply it by 64, .

.

Subtract the result from Equation 2

Substitute new equation for Equation 2


Forward elimination step 1 cont2

Forward Elimination: Step 1 (cont.)

Divide Equation 1 by 25 and

multiply it by 144, .

.

Subtract the result from Equation 3

Substitute new equation for Equation 3


Forward elimination step 21

Forward Elimination: Step 2

Divide Equation 2 by −4.8

and multiply it by −16.8,

.

.

Subtract the result from Equation 3

Substitute new equation for Equation 3


Finding the determinant

Finding the Determinant

After forward elimination

.


Summary

Summary

  • Forward Elimination

  • Back Substitution

  • Pitfalls

  • Improvements

  • Partial Pivoting

  • Determinant of a Matrix


Additional resources

Additional Resources

For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit

http://numericalmethods.eng.usf.edu/topics/gaussian_elimination.html


Gaussian elimination

THE END

http://numericalmethods.eng.usf.edu


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